考虑层序遍历
如果当前节点的左右子树任一为空,判断同层的下一个节点的左右子树是否都为空即可
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param root TreeNode类 * @return bool布尔型 */ int num = -1; public boolean isCompleteTree (TreeNode root) { if (root == null) { return true; } Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); while (size-- > 0) { TreeNode node = queue.poll(); if (node.left == null && node.right != null) { return false; } if (num > -1 && (node.left != null || node.right != null)) { return false; } if (node.left == null || node.right == null) { num = 0; } if (node.left != null) { queue.offer(node.left); } if (node.right != null) { queue.offer(node.right); } } } return true; } }